10.3 Instrument Aided Observations 83 the name of the selected star notify the examiner to check it. Deneb (Alpha Cygni) / Alfirk (Beta Cephei) / Algol (Beta Persei) / Capella (Alpha Aurigae) (b) Note down readings on dials on the telescope mount again. Estimate the “declination” and the “hour angle” of the star you have chosen. 10. The telescope for this task is aimed at open cluster NGC 6231. Ensure that 10mm Plössl eyepiece is put on the telescope. Estimate the diameter of the field of view of this telescope using the two angular distances shown on figure 10.11 as reference. Express your answer in arc minutes. (I12 - O02 - C) 11. Use figure 10.12 to estimate the magnitude of the missing star, shown as a cross, inside NGC 6231. Use magnitude of other stars as reference. (I12 - O03 - C) Note: To avoid confusion between decimal point and real stars, decimal points are not shown. So, magnitude 60 correspond to magnitude 6.0. Give your answer with a precision of 0.1 magnitude. Also note that 8.0 star in the chart is a variable. 12. You have to identify as many stars as possible in the field of a celestial photograph using the telescope and CCD provided. You can choose one out of the five recommended regions in the sky listed below. Then, point the telescope to the direction of the selected sky region. Take three photographs with different exposure times and record the images of the sky by the CCD camera. Save the observational data. Transfer the data to the printing facilities to print out the result. Ask the technical assistant for a help. Choose the best prints-out and use the image to identify the stars in the field of observations. Step by step process for this is given below. (I08 - O02 - D) (a) Choose only one out of the following five recommended regions to the directions of (marked by the following bright clusters): – M7: (α = 17h53m.3, δ = −34◦46′ .2) – M8: (α = 18h 04m.2, δ = −24◦ 22′ .0) – M20: (α = 18h02m.4, δ = −22◦58′ .7) – M21: (α = 18h04m.2, δ = −22◦29′ .5) – M23: (α = 17h 57m.0, δ = −18◦ 59′ .4) You may not change your choice. (b) Point the telescope to the chosen cluster using telescope controller. If necessary you may move the telescope slightly to get the best position in the frame of the CCD by checking the display of “CCDops” software.
84 Night Sky Observation Figure 10.11 – Image of NGC 6231 with angular separation
10.3 Instrument Aided Observations 85 Figure 10.12 – Image of NGC 6231 with magnitudes (c) Display the region in The Sky map software provided in the computer, to confirm that the telescope is pointed to the selected object in the sky. You may change field of view of the sky chart. (d) You may invert the background images into white color, as in the chart mode. Copy and paste the sky chart from The Sky into the answer page. Use “Ctrl-c” and “Ctrl-v” buttons from the keyboard, respectively. (e) Type the equatorial coordinates of the centre of that object in the answer page as indicated in The Sky. (f) Take three photographs of the chosen object by using the attached CCD camera and CCDops software, with various exposure times. Choose exposure time in the range between 1 and 120 seconds. Image is automatically subtracted with dark frame of the same exposure time. (g) You must invert the background images into white color. Copy and paste images from CCDops into the answer page. Use “Ctrl-c” and “Ctrl-v” buttons from the keyboard, respectively. (h) Save your answer page into hard disk in a Microsoft Word file
86 Night Sky Observation format. (i) Print your answer page which consists of the photographs and the corresponding sky chart. (j) Go to the identification room and bring with you the prints-out of the sky chart and the photographs. Ask some help from technical assistant, if necessary. (k) Choose the best out of the three printed images and identify as many objects as possible on it. (l) Use the assigned computer and The Sky software to identify objects. Type your identification to your answer page. (m) Type on the answer page the names (or catalogue number), RA, Dec, and magnitude of each identified star and put the sequential number on the photograph. Make sure that you list the stars on the answer page following the same order and number as on the photograph. (n) Estimate the limiting magnitude of the photograph you choose, empirically. 10.4 Planetarium Based Questions 1. Planetarium is projecting the sky of Bejing, on some day, just about 1 hour after the sunset. Estimate which month (use 1-12) it should be according to the displayed night sky. What is the age 1 of the moon (in 1-30 days)? (I10 - O03 - B) The co ordinates of Beijing: Longitude: 116◦ 48’E, Latitude: 40◦ 32’N. 2. Simulation of the Earth Sky (I11 - O01 - C) (a) In the sky projected, one can notice cresent moon, a nova and a comet. On the map of the sky (use figure 10.3), mark nova with a cross and label it as “N”. Mark the Moon with a Moon symbol and draw the shape and position of the comet. (b) In the table below, circle only those objects which are above the astronomical horizon. M20 - Triffid Nebula o Cet - Mira δ CMa - Wezen α Cyg - Deneb M57 - Ring Nebula β Per - Algol δ Cep - Alrediph α Boo - Arcturus M44 - Praesepe (Beehive Cluster) 1. Age of the moon is number of days since last new moon.
10.4 Planetarium Based Questions 87 (c) Coordinate grid will now be switched on. When it is visible, mark on the map the northern part of the local meridian (from the zenith to the horizon) and the ecliptic north pole (with a cross and marked “P”). (d) For the displayed sky, note down: geographical latitude of the observer (φ): Local Sidereal Time (θ): approximate time of year (by circling the calendar month): Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec. (e) Give the names of the objects, whose approximate horizontal coordinates are : azimuth A1 = 45◦ and altitude h1 = 58◦ : azimuth A2 = 278◦ and altitude h2 = 20◦ : (If you can, use Bayer designations, IAU abbreviations and Messier numbers or English or Latin names.) (f) Give the horizontal coordinates (azimuth, altitude) of: Sirius (α CMa): A3 = ; h3 = The Andromeda Galaxy (M31): A4 = ; h4 = (g) Give the equatorial coordinates of the star marked on the sky with a red arrow: α = ; δ = 3. Simulation of Martian sky: The sky projected now is as it appears for an observer standing at some distance from a martian base. The martian base is visible on the horizon. (I11 - O02 - C) (a) Give the areographic (Martian) latitude of the observer : ϕ = (b) Give the altitudes of upper (hu) and lower (hl) culmination of: Pollux (β Gem) : hu = ; hl = Deneb (α Cyg) : hu = ; hl = (c) Give the areocentric (Martian) declination of: Regulus (α Leo) δ= Toliman (α Cen) δ= (d) Sketch diagrams to illustrate your working in the two sub-questions 3b and 3c above:
88 Night Sky Observation (e) On the map of the sky, mark the Martian celestial North Pole with a cross and label it as “M”. (f) Estimate the azimuth (measured from the South) of the observer as seen from the Martian base visible on the horizon. (g) Estimate the location of the base on Mars and circle the appropriate description: a. near the northern Tropic circle b. near the Equator c. near the northern Arctic circle d. near the North Pole (h) The time axis below shows the Martian year and the seasons in the northern hemisphere. Mark the date corresponding to the sky on the axis. Autumn Winter t Spring Summer
Chapter 11 Solutions: Celestial Mechanics 11.1 Theory 1. (perihelion distance) + (aphelion distance) = (major axis of the orbit). Thus, semi-major axis, a = 31.5 + 0.5 2 = 16 A. U. According to Kepler’s third law, we have T 2 ∝ a 3 (11.1) T 2 = (constant)a 3 This constant is 1(year) 2 (A.U.) 3 , for our Solar system, when T is measured in years and a in A.U.s. This is a useful trick to remember for comparision of periods of the solar system object. For this comet we have, T 2 = 163 = (4)6 = 4 3 2 T = 64.0 years (11.2) 2. This comet has period of 64 years and semi-major axis measuring 16 A. U. By Kepler’s second law, the comet will sweep equal area in equal time. ∴ Area per year = Area of elliptical orbit T = πab T (11.3) The perihelion distance is given by a(1 − e) = 0.5 A. U., where e is the
90 Solutions: Celestial Mechanics eccentricity. b 2 = a 2 (1 − e 2 ) (11.4) = a 2 1 − 1 − 0.5 a 2 ! = a 2 − (a − 0.5)2 = 162 − 15.5 2 b = 3.97A.U. (11.5) ∴ Area per year = π × 16 × 3.97 64 = 3.1 (A.U.) 2 /year (11.6) 3. As the comet is entering inner solar system, its perihelion distance will be of the order of 10 A.U. or less, which is very small as compared to aphelion distance. a = dperi + dap 2 (11.7) ≈ dap 2 = 35000 2 = 17500AU (11.8) T = a 1.5 (11.9) = 2.3 × 106 years (11.10) tjourney = T 2 (11.11) = 1.2 × 106 years. (11.12) 4. Let us assume that A normal man can jump vertically up to 50 cm on the Earth. This number is purely a rough estimate and one can choose
11.1 Theory 91 any number between 10 cm to 50 cm. vjump = q 2gh (11.13) vescape = s 2GM R (11.14) ∴ 2gh = 2GM R = 2G R 4πρR3 3 R = s 3gh 4πρG (11.15) = s 3 × 9.81 × 0.5 4π × 5515 × 6.67 × 10−11 = 1.78 × 103m (11.16) R ≈ 2km (11.17) 5. The mass of the asteroid is m = 4 3 πr3 ρ (11.18) = 4 3 π(1.1 × 103 ) 3 × 2.2 × 103 = 1.23 × 1013Kg (11.19) Since,mastronaut ≪ m it can be safely ignored. Then, critical velocity is, vcrit = s Gm r (11.20) = s 6.6726 × 10−11 × 1.23 × 1013 1100 = 0.863 Km/s (11.21) If the velocity of the astronaut is greater then vcrit, he will start orbiting the satellite. That means he will not be able to “walk” on the satellite. If the astronaut wants to complete a circle along the equator of the asteroid on foot within 2.2 hours, v2 = 2π × 1100 2.2 × 3600 (11.22) = 0.873Km/s (11.23) As v2 > vcrit, the astronaut will NOT be able to complete the circle on foot within stipulated time.
92 Solutions: Celestial Mechanics 6. As the stars are all part of the cluster, the r.m.s. velocity must be smaller than the escape velocity. As an rough estimate, one can take vesc = √ 2vrms (11.24) However, this is just a convenient assumption and one is free to make similar other assumptions. 1 2 Mv2 esc = 1 2 M GM R (11.25) v 2 esc = 2GM R ∴ v 2 rms = GM R ∴ M = Rv2 rms G (11.26) = 20 × 3.0856 × 1016 × 9 × 106 6.672 × 10−11 = 8.3 × 1034 Kg (11.27) M = 4.2 × 104M⊙ (11.28) 7. The escape velocity of object on the edge of the cluster is given by, ve = s 2GMcl Rcl (11.29) ∴ Mcl = NM⊙ = Rclv 2 e 2G ∴ N = Rclv 2 e 2GM⊙ (11.30) = 20 × 3.0856 × 1016(6 × 103 ) 2 2 × 6.6726 × 10−11 × 1.9891 × 1030 N ≈ 8.4 × 104 (11.31)
11.1 Theory 93 8. To determine type of orbit, we find total energy of the spacecraft. ET ot = KE + P E = 1 2 mv2 − GMm R (11.32) = m (17062)2 2 − 6.6726 × 10−11 × 1.9891 × 1030 116.406 × 1.4960 × 1011 ! = m 145.556 × 106 − 7.62158 × 106 ET ot = 1.3793 × 108m (11.33) ET ot > 0 (11.34) Hence the orbit is Hyperbolic. The magnitude of the Sun will be given by, m1 − m2 = −2.5 log f1 f2 ! (11.35) m1 − m2 = −2.5 log d 2 2 d 2 1 ! m1 = m2 + 5 log d1 d2 ! (11.36) = −26.72 + 5 log 116.406 1 m1 = −16.39 (11.37) 9. Ratio of average densities of the Sun and the Earth will be given by ρ⊕ ρ⊙ = M⊕ (R⊕) 3 (R⊙) 3 M⊙ (11.38) T⊕ = 2π s a 3 GMS (11.39) Let θ⊙ be the angular diameter of Sun in radians. θ⊙ = 2R⊙ a (11.40) ⇒ a = 2R⊙ θ⊙ T 2 = 4π 2 × 8R⊙ 3 GM⊙θ 3 ⊙ ⇒ R3 ⊙ M⊙ = GT2 θ 3 ⊙ 32π 2 (11.41)
94 Solutions: Celestial Mechanics Using expression for arc legnth of a circle and definition of g, g = GM⊕ (R⊕) 2 ⇒ M⊕ (R⊕) 2 = g G (11.42) R⊕ϕ = S 1 R⊕ = ϕ S (11.43) Combining equations 11.41, 11.42 and 11.43, we get ρ⊕ ρ⊙ = M⊕ (R⊕) 2 1 R⊕ (R⊙) 3 M⊙ = g G ϕ S GT2 θ 3 ⊙ 32π 2 = gϕθ3 ⊙T 2 32π 2S (11.44) = g × π 180 × π 3603 × T 2 32π 2S = gT2π 2 32 × 180 × (360)3S (11.45) = 9.81 × (3.1557 × 107 ) 2 π 2 32 × 180 × (360)3 ρ⊕ ρ⊙ = 3.23 (11.46) 10. Here we have to balance the gravitational attraction of the Sun with the radiation pressure. The force of radiation pressure is simply solar radiation flux received by an object divided by the speed of light. Assuming that the dust grains completely abosorb all the radiation incident on them, Force due to gravitational attraction = GMm r 2 Force due to radiation process = L 4πr2 πD2 4c [Dust grains completely absorbs
11.1 Theory 95 all radiation] LD2 16r 2c For equilibrium, FRP = L⊙ 4πa2 ⊕ πr2 dust c (11.47) Fgrav = FRP (11.48) ∴ GM⊙m a 2 ⊕ = L⊙r 2 dust 4a 2 ⊕c m = L⊙r 2 dust 4GM⊙c 4π 3 r 3 dustρ = L⊙r 2 dust 4GM⊙c ∴ rdust = 3L⊙ 16πρGM⊙c (11.49) = 3 × 3.826 × 1026 16π × 103 × 6.6726 × 10−11 × 1.9891 × 10303 × 108 rdust ≈ 0.6µm (11.50) 11. τ = Iα (11.51) = 2 5 M⊕R 2 ⊕α ωi = ωf + αt (11.52) = 2π daysidereal + τ 2 5M⊕R2 ⊕ × 6 × 108 × (1 year) = 2π 23h56m4 s .1 + 5 × 6 × 1016 × 6 × 108 × 3.1557 × 107 2 × 5.9736 × 1024 × (6.3708 × 106 ) 2 = 7.292 × 10−5 + 1.171 × 10−5 ωi = 8.463 × 10−5 rad/s (11.53) daysi = 1 sidereal year Ti = 1 sidereal year × ωi 2π (11.54) = 3.15 × 107 × 8.463 × 10−5 2π = 424.3 days daysi ≈ 420 days (11.55)
96 Solutions: Celestial Mechanics 12. The mechanical energy is conserved and has the form E = p 2 2m − GMm r = p 2 2m − GMmu (11.56) EA = EB (11.57) p 2 A 2m − GMmuA = p 2 B 2m − GMmuB (11.58) p 2 A 2m − p 2 B 2m = GMmuA − GMmuB (11.59) (p 2 A − p 2 B) 2m = GMm(uA − uB) (11.60) ∴ m = vuut (p 2 A − p 2 B) 2GM(uA − uB) (11.61) m = vuut (5.2 × 107 ) 2 − (1.94 × 109 ) 2 2 × 6.67 × 10−11 × 5.97 × 1024 × (5.15 − 194.17) × 10−8 (11.62) = 5.0 × 104 kg (11.63) E = EA = p 2 A 2m − GMmuA (11.64) E = (5.2 × 107 ) 2 2 × 50000 − 6.67 × 10−11 × 5.97 × 1024 × 5 × 104 × 5.15 × 10−8 (11.65) = −1.0 × 1012J (11.66) To sketch the curve, we use the equation E = p 2 2m − GMmu (11.67) → u = p 2 2GMm2 − E GMm (11.68)
11.1 Theory 97 This is a parabolic curve. 13. We can solve the question in two ways: (I) Algebraic Solution As the spacecraft is stationary at (0,0), let y = mx be the line of sight of the telescope. Then, the intersection of the line of sight and the circle is x 2 + (mx) 2 − 10x − 8(mx) + 40 = 0 (11.69) (1 + m2 )x 2 − (10 + 8m)x + 40 = 0 (11.70) The equation will have solution if its discriminant (D = b 2 −4ac) is greater than or equal to zero. D = b 2 − 4ac ≥ 0 (11.71) (−10 − 8m) 2 − 4(1 + m2 )40 ≥ 0 (11.72) 64m2 + 160m + 100 − 160m2 − 160 ≥ 0 (11.73) −96m2 + 160m − 60 ≥ 0 (11.74) −24m2 + 40m − 15 ≥ 0 (11.75) The solution of the inequality is (10 − √ 10) 12 ≤ m ≤ (10 + √ 10) 12 (11.76) Therefore these extreme values will be values of tan φ. (II) Geometric Solution x 2 + y 2 − 10x − 8y + 40 = 0 (11.77) x 2 − 10x + 25 + y 2 − 8y + 16 − 1 = 0 (11.78) (x − 5)2 + (y − 4)2 = 1 (11.79) The centre of the circle is (5, 4) and the radius is 1 unit. Now, we should find tangents from origin to this circle to give
98 Solutions: Celestial Mechanics minimum and maximum angles of elevation. For this purpose, we define angle α as the angle made by the center of the disk with x-axis and angle β as angle subtended by radius of the disk at the origin. tan α = 4 5 (11.80) tan β = 1 √ 5 2 + 42 − 1 = 1 √ 40 = 1 2 √ 10 (11.81) tan φmin = tan(α − β) (11.82) = tan α − tan β 1 + tan α tan β (11.83) = 4 5 − 1 2 √ 10 1 + 4 10√ 10 (11.84) = (8√ 10 − 5) (10√ 10 + 4) (11.85) = (8√ 10 − 5)(10√ 10 − 4) (10√ 10 + 4)(10√ 10 − 4) (11.86) = (800 − 50√ 10 − 32√ 10 + 20) (1000 − 16) (11.87) = 82(10 − √ 10) 82 × 12 (11.88) = (10 − √ 10) 12 (11.89) tan φmax = tan(α + β) (11.90) = tan α + tan β 1 − tan α tan β (11.91) = (10 + √ 10) 12 (11.92) 14. For the cloud, let O be the centre and A be some point on the surface.
11.1 Theory 99 Thus, the object is traveling from O to A. Let Vi be gravitational potential at point i. If the particle was free-falling, then its velocity at the surface will be be zero. KEA + P EA = KEO + P EO (11.93) 0 + VA = GMm 2R + VO (11.94) ∴ VO = VA − GMm 2R = − GMm R − GMm 2R VO = − 3GMm 2R (11.95) To escape, total energy of the particle should be zero. If VO = Vesc KEO + P EO = 0 (11.96) 1 2 V 2 esc + VO = 0 (11.97) 1 2 V 2 esc = 3GM 2R ∴ Vesc = s 3GM R (11.98) 15. In this figure, D is the centre of the Moon. The moon’s gravitational acceleration at A and the centre of the Earth C are, −→gA = − GMm r 3 A −→rA (11.99) −→gC = − GMm r 3 C −→rC = − GMm r 3 −→rC (11.100)
100 Solutions: Celestial Mechanics Figure 11.1 – High altitude projectile Both these vectors point towards the Moon. From the figure, rA = AD and rC = CD. The Moon’s net gravitational acceleration felt by test mass at A is, −→ g ′ A = −→gA − −→gC = − GMm r 3 A −→rA − GMm r 3 −→rC ! (11.101) From the figure, −→rA = −→rC + −→R. Thus, −→ g ′ A = − GMm r 3 A −→R − GMm r 3 A − GMm r 3 ! −→rC (11.102) Now, first term of the R.H.S. is purely radial. Thus, for horizontal acceleration of the sea water at A, we have to only take horizontal component of the second term. Distance rA can be found by cosine rule. −−−−→ g ′ A(hori) = GMm r 3 − GMm r 3 A ! r sin φ (11.103) = 1 r 3 − 1 (r 2 + R2 − 2Rr cos φ) 3/2 ! GMmr sin φ (11.104) 16. In the figure 11.1, the projectile is launched from point A on the surface of the earth, it reaches highest point at H and then hits the Earth surface back at B. The principal focus of the orbit is at the centre of the earth (O). Point M is on the surface of the earth along the major axis (LOH). Line AB cuts the major axis at C and O’ is the other focus of the ellipse. Tangent at point A meets the major axis between M and O’. One should remember that when the projectile goes from the surface to high altitude, the gravitational acceleration will be variable along the
11.1 Theory 101 path. Hence approximated relations, which assume constant g, cannot be used. (a) Total energy of the projectile is E = 1 2 mv2 0 − GM⊕m R⊕ (11.105) = GM⊕m 2R⊕ − GM⊕m R⊕ (11.106) = − GM⊕m 2R⊕ < 0 (11.107) As the total energy is less than 0, it means that orbit might be ellipse or circle. As θ > 0, the orbit is an ellipse. Total energy for an ellipse is E = − GMm 2a (11.108) Comparing equation 11.107 and 11.108, we conclude a = R (11.109) (b) Let b be the semi-minor axis of the orbit. b = a √ 1 − e 2 (11.110) = R⊕ √ 1 − e 2 (11.111) b = R⊕ sin π 3 = √ 3 2 R⊕ (11.112) ∴ √ 1 − e 2 = √ 3 2 1 − e 2 = 3 4 e 2 = 1 4 ∴ e = 0.5 (11.113) rmax = R⊕(1 + e) (11.114) = 1.5R⊕ (11.115) ∴ hmax = HM = HO − OM (11.116) = rmax − R⊕ h = 0.5R⊕ (11.117)
102 Solutions: Celestial Mechanics (c) As there is no air drag or other dissipitive force, paths AHB will be symmetric about the major axis. As ∡COA = 60◦ , we can conclude, ∡BOA = 120◦ . Thus range from A to B along the surface will be, R = z}|{ AB = 120 360 × 2πR⊕ (11.118) R = 2π 3 R⊕ (11.119) (d) As already shown in equation 11.113, e = 0.5 (e) We can show that C is actually the centre of the ellipse by showing that OC = ae. l(OC) = l(OC) sin(∡OAC) (11.120) = a sin 30◦ l(OC) = 0.5a = ae (11.121) ∴ b = a √ 1 − e 2 (11.122) = a q 1 − (0.5)2 = a s 1 − 1 4 = a s 3 4 b = √ 3a 2 (11.123) Now by Kepler’s second law, time spent to cover particular part of the orbit will be proportional to the area swept by the radius
11.1 Theory 103 vector. Thus, we want to find Area(OAHBO). Area(OAHBO) = Area(CAHBC) + Area(△AOB) (11.124) = πab 2 + 1 2 × l(OC) × l(AB) (11.125) = πab 2 + 1 2 × 0.5a × 2b = πab 2 + ab 2 Area(OAHBO) = (π + 1)ab 2 (11.126) T = vuut 4π 2a 3 GM⊕ (11.127) ∴ t T = Area(OAHBO) Area(LAHBL) (11.128) ∴ t = (π + 1)ab 2πab × 2π vuut a 3 GM⊕ (11.129) = (π + 1) vuut a 3 GM⊕ (11.130) = (π + 1)s (6.3708 × 106 ) 3 6.672 × 10−11 × 5.9736 × 1024 = 3336 sec (11.131) t = 55m36s (11.132) 17. As the orbit is parabolic, Esat = −GM⊙ rsat + 1 2 v 2 sat msat = 0 (11.133) ∴ vsat = s 2GM⊙ rsat (11.134) (a) Angular momentum of the satellite at the launch is, L = msatvsat,⊕a⊕ = msatq 2GM⊙a⊕ (11.135) Let vT is the tangential velocity of the satellite at the Mars’ orbit.
104 Solutions: Celestial Mechanics Conserving the angular momentum, msatq 2GM⊙a⊕ = msatvT am ∴ vT = √ 2GM⊙a⊕ am (11.136) cos ψ = vT v = √ 2GM⊙a⊕ am s am 2GM⊙ (11.137) = s a⊕ am = s 1 1.52 = 0.8111 (11.138) ∴ ψ = 35.80◦ (11.139) (b) If the Mars is very close to the satellite, then orbital velocity of the Mars will be parallel to the tangential velocity of the satellite. −→vrel = −−→vrad + −→vT − −→vm (11.140) = vradrˆ + (vT − vm) ˆθ (11.141) vm = s GM⊙ am (11.142) = s 6.6726 × 10−11 × 1.9891 × 1030 1.52 × 1.496 × 1011 vm = 24.16 km/s (11.143) v = s 2GM⊙ rm = √ 2vm (11.144) = 34.17 km/s (11.145) vT = v cos ψ = 34.17 × 0.8111 (11.146) = 27.71 km/s (11.147) vrad = v sin ψ (11.148) = 19.99 km/s (11.149) ∴ −→vrel = 19.99ˆr + 3.55ˆθ (11.150) | −→vrel| = 20.30 km/s (11.151) θ = tan−1 3.55 19.99 (11.152) = 10.07◦ (11.153) Thus the angle is about 10◦ eastwards. As the satellite has been launched in the direction of the Earth’s motion and given the
11.1 Theory 105 fact that the Mars and the Earth have same sense of revolution around the Sun, −→vT and −→vm are in the same direction and hence the magnitudes have to be subtracted. 18. In this question the trick is to realise that the two angles α and β are not in the same plane. In other words, the impulse given (∆−→p ) is not in the plane of the orbit of the satellite. If the orbit is in x-y plane, we can say, −→ p ′ = −→p + ∆−→p (11.154) p ′ tangential = p + p cos α sin β (11.155) p ′ radial = p cos β (11.156) p ′ z = p sin α sin β (11.157) We note that if cos β 6= 0 then, the point of explosion will no longer remain either perigee or apogee point. Thus, β = 90◦ or 270◦ (11.158) Thus, we get p ′ = q p 2 (1 + cos α) 2 + p 2 sin2 α (11.159) = p q 1 + cos2 α + 2 cosα + sin2 α p ′ = p q 2(1 + cos α) = p q 4 cos2 (α/2) p ′ = 2p cos(α/2) (11.160) Now, for orbit to remain circular, we need p ′ = p. For parabola we need, p ′ = √ 2p. For ellipse, we need p ′ to be smaller than the value for parabola and for hyperbola we need values greater than that of parabola. Thus, Type α hyperbola perigee = initial point [0◦ : 90◦ ) or (270◦ : 360◦ ] parabola perigee = initial point 90◦ or 270◦ ellipse perigee = initial point (90◦ : 120◦ ) or (240◦ : 270◦ ) circle 120◦ or 240◦ ellipse apogee = initial point [120◦ : 180◦ ) or (180◦ : 240◦ ] At exactly α = 180◦ , we get p ′ = 0 and the satellite will fall on the earth. This also assumes the Earth to be a point object. However, in reality, for angles close to α = 180◦ , the orbit will pass so close to the centre of the Earth, the satellite will hit the Earth surface.
106 Solutions: Celestial Mechanics 11.2 Data Analysis 1. It was expected that students should pay attention the motion of the moons. By Kepler’s third law, shorter the orbital radius, shorter the period. Students were expected to know the sequence of moons (from closer to Jupiter to farther) as Io, Europa, Ganymede and Callisto.
Chapter 12 Solutions: Celestial Coordinate Systems 12.1 Theory 1. α = Diameter of Earth Distance to Moon (12.1) = D⊕ de−m radians = D⊕ de−m × 180 × 60 π arcminutes (12.2) = 12742 378000 × 180 × 60 π arcminutes (12.3) = 116′ = 116 15 minutes of RA (12.4) Thus, the difference in apparent right ascension is 7.73 minutes. 2. As the solar day is measured from noon to noon, it is slightly longer than the rotation period of the Earth (sidereal day). Over the course of one solar year, this small time is accumulated to one additional sidereal day. Thus, in exactly half a year, it would have accumulated to 0.5 extra days. Thus, there will be 183.5 sidereal days from the noon of 1 July to the noon of 31 December. Mathematically, the sidereal day is 3m 56s shorter than the solar day. ∴ 183 solar days = 183d + 183 × (3m56s ) = 183d 11h 50m48s ≈ 183.5 days (12.5)
108 Solutions: Celestial Coordinate Systems 3. For condition (a), we have to ensure that the ray of light, grazing the outer edge of the Tabeshband, should hit lower edge of the window. For condition (b), we have to ensure that the ray of light, grazing the outer edge of the Tabeshband, should reach point A as in figure 2.1. The zenith angle of the Sun at the summer and winter solstice will be given by, zs = φ − 23.5 ◦ = 12.5 ◦ (12.6) zw = φ + 23.5 ◦ = 59.5 ◦ (12.7) From the figure, (right) at the summer solstice we have tan(zs) = x h (12.8) = tan 12.5 ◦ = 0.222 (12.9) And in the winter solstice, tan(zw) = D + x H (12.10) = tan 59.5 ◦ = 1.70 (12.11) ∴ x = 1.70H − D x = 0.593m (12.12) h = x/0.222 (12.13) h = 2.67m (12.14) 4. To find the minimum declination, we have to consider various effects. Firstly, we have to note the latitude of the observer. Then we note that as the observer is at a high altitude, he/she will be able to see some part of the sky below the standard horizon. This is known as horizon depression (θ). Thirdly, atmospheric refraction at the horizon (α) is about 34’. cos θ = R R + h = 6370.8 6370.8 + 5.6 (12.15) θ = 2.40◦ = 2◦ 24′ (12.16) ∴ δ = 90◦ − φ − θ − α (12.17) = 90◦ − (35◦ 57′ + 2◦ 24′ + 34′ ) = 90◦ − 38◦ 55′ ∴ δ = 51◦ 5 ′N (12.18)
12.1 Theory 109 5. Recall coordinate transformation relations between equatorial and horizontal coordinates. For the special case of rising or setting of an astronomical object, the relation is reduced to a simplified form cos H = − tan δ tan φ (12.19) For 21st December, δ⊙ = −23.5 ◦ and question states φObserver = 42.5 ◦ . Thus, H = cos−1 (− tan(−23.5) tan 42.5) = cos−1 (0.4348 × 0.9163) = cos−1 (0.3984) H ≈ 4 h 26m5 s (12.20) Thus, equation 12.20 shows that the Sun rises 4 hours 26 minutes before local noon or at 7:34am local time. Next we notice that the longitude of observer is λObserver = 71◦ E. As local time changes by 4 minute for every degree of longitude, his local time must be 4 hours and 44 minutes ahead of GMT. But the problem states that his civil time is 5 hours after GMT. This means his civil time is 16 minutes behind of his local time. Applying this correction, the Sunrise time as per the observer’s watch will be, t = 7 : 34 − 0 : 16 = 7 : 18am (12.21) EA: Correction for atmospheric refraction is roughly 34’ and angular radius of solar disk can be upto 16’. As a combined effect of these two factors, the top edge of the Sun becomes visible when the centre of the Sun is about 50’ below horizon. In the figure, ABC is the horizon line. A is the position where the centre of the Sun will rise. B is the place where top edge of the Sun becomes visible when the centre of the Sun is at S. Thus, l(BS) = 50′ . It can be seen that ∡BCS = ∡BSA = φ (12.22) ∴ AS = l(BS) cos φ = 50′ cos 42.5 ◦ ≈ 67′ .8 (12.23)
110 Solutions: Celestial Coordinate Systems As this is a small length, we approximate the Sun’s speed as 4 minutes per degree. Thus, the Sun will rise about 4.5 minutes earlier due to these two factors. The equation of time shows that the Sun will be about 2 minutes (1.7 minutes to be precise) ahead on 21st December. Thus total error introduced by all these effects is about 6 minutes. 6. When ever the Sun is towards the South of prime vertical (i.e. the great circle through East, West and Zenith), the sundial will not work. (i) When the Sun is south of equator i.e. −23◦26′ < δ < 0 , it will set south of west and thus across the prime vertical. Thus for this period from September 23 to March 21, there will be at least some part of the day for which the Sun will not cast any shadow on the sundial. This corresponds to the spring and summer season of southern hemisphere. (ii) It will not work the whole day, if the Sun is south of prime vertical for the whole day i.e. it should cross the meridian on the south of zenith. This will happen when declination of the Sun is less than latitude of the place. −ǫ < δ < φ (12.24) −23◦ 27′ < δ < −22◦ 54′ (12.25) λ = sin−1 sin δ sin ǫ ! (12.26) = sin−1 (0.9778) 77.91◦ < λ < 102.09◦ (12.27) This corresponds to 11.9 days on the either side of winter solstice (22nd Dec. 7. Note that the date of observation is (June 19) very close to summer solstice. Thus, δ ≈ +23. ◦ .5, & α ≈ 6 h At full moon phase, the Moon is nearly diametrically opposite to the Sun along the ecliptic. The diametrically opposite position will be, δ ≈ −23. ◦ .5, & α ≈ 18h We further note that the Moon can be upto 5◦ .25 above or below the ecliptic. Thus, δ can be between −18◦ .25 and −28◦ .75 depending on
12.1 Theory 111 the position of node. cos(H) = − tan δ tan φ (12.28) ∴ H1 = cos−1 (− tan δ tan φ) (12.29) = cos−1 (− tan(−18.5) tan(−6 ◦ 49′ ) (12.30) = 92◦ .3 = 6h 9 m (12.31) ∴ H1 = cos−1 (− tan δ tan φ) (12.32) = cos−1 (− tan(−28.75) tan(−6 ◦ 49′ ) (12.33) = 93◦ .8 = 6h 15m (12.34) As a first approximation, the duration will be given by 2H i.e. Its extreme values can be from 12 hours 18 minutes and 12 hours 30 minutes. However, we have to include two important corrections to this value. Firstly, due to the lunar orbital motion, the Moon is moving backwards by slightly less than 53 minutes per day. Thus, in 12.5 hours, it will move backwards by nearly 27 minutes and will be visible for 27 more minutes. Secondly, as the refraction near horizon is nearly half a degree, the moon will be seen longer by 2 minutes each at rising and setting. Thus, total correction will be for 31 minutes. Hence, extreme possible values are 12 hours 49 minutes and 13 hours 1 minute. One may note that some of the data given in this problem like longitude, civil time and elevation is not used in the solution. It is expected that student solving such problems chooses data judiciously and not get confused by excess data. 8. Since FOV is very small, motion of Vega across the FOV can be approximated by a straight line. We will calculate angle of rotation of the sky near Vega’s location. We must remember to include the declination effect. ∴ θ = 360◦ × t Tsidereal × cos δ (12.35) = 360◦ × 5.3 (23 × 60 + 56) × cos 39 θ ≈ 62′ (12.36) 9. Recall that hour angle of a star, its R.A. and local sidereal time are related by α = t − H (12.37)
112 Solutions: Celestial Coordinate Systems Ν Ζ S E P W C G Figure 12.1 – Galactic plane in horizontal coordinates At lower culmination, hour angle is 12 hours. Hence, the sidereal time at that moment will be t = α + H = 14h 51m + 12h 00m t = 2h 51m (12.38) This will be your lowermost marker on the dial. Next we should ensure that scale is marked “anti-clockwise” as that will be direction of motion of the stars. On the UT side of the card, we again note that lowermost point will correspond to 03:15 for the given date and scale will again be anti-clockwise. Now as sidereal day is shorter than the solar day by 4 minutes, the clock will need second dial to calibrate for this effect. This can be done easily by calibrating inner dial with dates. 10. As the current local sidereal time (LST) is exactly 12 hours away from RA of the northern galactic pole (NGP), it will be at the Lower culmination. In figure 12.1, P is the NCP, G is NGP and line EW marks the galactic equator. Plane NWS shows the horizon where letters represent directions. In the equation 2.1, ǫ is basically the angle between the poles of the two co-ordinates. Thus, it will be replaced by angle between ZG. Also, from the figure, it is clear that the altitude of galactic equator will be maximum at A = 0◦ (North) and minimum at A = 180◦ (South).
12.1 Theory 113 Note that A = 90◦ at the point where the galactic equator meets the horizon. As compared to this, R.A. is zero at the meeting point of ecliptic. Also as the NGP is below horizon, the directions of increasing coordinates is opposite. Thus, sin α should be replaced by − cos A. ZG = P G + P Z (12.39) = 90◦ − δG + 90 − φ = 180◦ − (δG + φ) (12.40) h = tan−1 [− cos A tan(180 − (δG + φ))] = tan−1 [cos A tan(δG + φ)] (12.41) φ + δG = 49◦ 39′ + 27◦ 08′ = 74◦ .7 (12.42) h = tan−1 [cos A tan 74◦ .7] (12.43) 11. One should always remember that the precession does not change ecliptic latitudes of the stars. At North Hemisphere, declination of the non-rising stars δ < ϕ − 90◦ , where ϕ stands for latitude, and declination of the non-setting stars δ > 90◦ − ϕ. So in Krakow, δnon−rise < −39.9 ◦ (12.44) δnon−set > 39.9 ◦ (12.45) Stars with intermediate declinations will rise and set. As an appproximate solution, we notice that both the stars are reasonably close to 6h R.A. Let us assume their R.A. to be exactly 6 hours. Currently at this R.A., declination of ecliptic (δec) is +23.5◦ , which is highest possible. If it changes to lowest possible value, i.e. δec = −23.5 ◦ , then, declination of both the stars will decrease by 47◦ . ∴ δSirius ≈ −63◦ (12.46) δCanopus ≈ −80◦ (12.47) It is not possible for declination of canopus to be much higher than its current declination. So it will never be possible for it to rise at Krakow. However, the new declination of Sirius is well inside never-rising region of the sky for Krakow. So it can become a never-rising star. One can solve this more rigourously using the spherical trigonometric formula given below. However, as the new declinations by approximate method are far away from boundary of never-rising region, such detailed treatment is not necessary sin β = sin δ cosǫ − cos δ sin ǫsin α (12.48)
114 Solutions: Celestial Coordinate Systems 12. Sunny solstice in southern hemisphere is December 23. On that day, declination of the Sun is δ = −23.45◦ . Thus, altitude of the Sun at this island on that noon will be, a0 = 90◦ + ϕ − δ (12.49) = 90◦ − 66.55◦ − (−23.45◦ ) a0 = 46.9 ◦ (12.50) At the Sunset, the centre of the Sun will just touch horizon and then start rising again. i.e. the lowest point of the declination circle along which the Sun travels on that day is just touching the horizon. The locus of tip of the shadow will be symmetric on the either side of the prime meridian (North - Zenith - South). Thus, as shown in the figure, the equation of the locus will be that of a parabola, with origin at the focus. Let us take a generalised parabola equation and find its coefficients. y = px2 + qx + r (12.51) Firstly, as the focus of the parabola is at the origin, q will be zero. We will take coordinates at noon as (x0, y0), and the coordinates when the Sun is at prime vertical (East - Zenith - West), say pre-noon, be (x1, y1). To find the second set, we have to find altitude of the Sun when its when it crosses prime vertical. We use the fact that the plane declination circle is inclined to the horizon at ǫ degrees. Let us say R is radius of the imaginary celestial sphere and z ′ be the projected height
12.1 Theory 115 of the Sun on z-axis. Therefore, x0 = 0 (12.52) y1 = 0 (12.53) y0 = r = hChristtana0 (12.54) = 39.60 × tan 46.9 ∴ r = −37.06m (12.55) z ′ = R sin a1 (12.56) tan ǫ = z ′ R = sin a1 (12.57) ∴ a1 = sin−1 (tan ǫ) = 25◦ 42′ (12.58) tan a1 = hChrist x1 (12.59) ∴ x1 = s −r p = hChrist tan a1 (12.60) = 82.26 ∴ p = −r x 2 1 (12.61) p = 0.0063 (12.62) ∴ y = 0.0063x 2 − 37.06 (12.63) 13. It is imporant to understand what is asked in the question. It is as if you are standing inside a rotating, huge sphere and various stars are attached to the sphere. Now one of these stars at the given location comes loose at the given time and flies off tangentially to infinity. We would like to know final apparent coordinates of this star and also time required for him to travel certain distance. Let R be the radius of the Celestial sphere and (x, y, z) a coordinate system with origin at the observer, z is the down-up axis, y is the eastwest axis and x is the south-north axis. At the superior culmination, we have, a = ϕ + δ = 7.41◦ + 49.35◦ (12.64) a = 56.76◦ (12.65) −→v = υ0yˆ = 2πRy /day ˆ (12.66) −→r0 = R(− cos axˆ + sin azˆ) (12.67) Now the object flies off at this stage. At any point in future, its location
116 Solutions: Celestial Coordinate Systems will be given by, −−→ r(t) = −→r0 + −→v t (12.68) a(t) = tan−1 z √ x 2 + y 2 ! (12.69) When t → ∞, the x and z coordinates of the star remain constant, while y → ∞ and we have a = tan−1 z ∞ = tan−1 0 = 0◦ (12.70) Similarly,the final azimuth is the West direction (A = 90◦ orA = 270◦ , depending on the system used). For the magnitude, m0 − m6 = −2.5 log F0 F6 = −5 log R6 R0 (12.71) R6 = 10 (6−0.45) 5 R R6 = 12.88R0 (12.72) | −−→ r(t)| = q x 2 + y 2 + z 2 (12.73) = q x 2 0 + (2πRt) 2 + z 2 0 = √ R2 + 4π 2R2 t 2 = R √ 1 + 4π 2 t 2 = 12.88R (12.74) ∴ 12.882 = 1 + 4π 2 t 2 t = s 12.882 − 1 4π 2 t = 2.044days (12.75) 14. As seen from an inertial reference frame outside the Earth, the observer must describe a circle parallel to the ecliptic, with constant speed, in a period of one sidereal day. Since the ecliptic poles will now be at zero latitude, this circle must be a great circle. But in the reference frame of the Earth, the total motion will be the motion in inertial frame (as described above) plus motion to compensate for the rotation of the Earth. In other words, the displacement of the observer is the vectorial addition of rotation through some angle ϕ along ecliptic and rotation through same angle ϕ around the Earth’s rotation axis. But this is exactly the same construction which would describe analemma of the Sun, if the Earth would have been in exact
12.2 Data Analysis 117 circular orbit around the Sun. So the resultant path of the observer will appear like an analemma for circular orbit (symmetrical figure of 8). It must be – closed curve. – should be bound by latitudes = ±ǫ. – it should be symmetric around the the equator as well as a chosen longitude. – It should have the shape of 8. As the observer starts on the southern hemisphere, he will cross the equator for the first time in the direction south-north. Its velocity relative to an inertial referential will have intensity υ = 2πR T and azimuth (the angle with the north-south direction) A1 = 90◦ −ǫ. The velocity of the Earth’s surface will have the same intensity and azimuth A2 = 90◦ . Subtracting the vectors, υcombined = υ q 2(1 − cosǫ) (12.76) Acombined = − ǫ 2 = −11.72◦ (12.77) 12.2 Data Analysis 1. This is an interesting problem as there is a lot of data but only a small part of it will be actually used to solve the question. Thus, key part of the solution is to first understand which data should be picked up from the table. (a) We start by around the dates of solstices and equinoxes, position of object A are as follows: R.A. Dec. h m s ◦ ’ ” Mar 21 23 59 51.47 - 0 1 1.44 Mar 22 0 3 30.11 + 0 22 40.3 Jun 21 5 56 41.18 + 23 26 13.51 Jun 22 6 0 50.78 + 23 26 20.99 Sep 23 11 58 23.75 + 0 10 27.2 Sep 24 12 1 59.32 - 0 12 54.79 Dec 22 17 58 42.17 - 23 26 27 Dec 23 18 3 8.7 - 23 26 21.28 This matches exactly with expected positions of the Sun on these dates. Thus, object A is the Sun.
118 Solutions: Celestial Coordinate Systems Now, we see how other objects move with respect to the Sun. If we scan through the entire table, we notice that the difference between R.A. of the Sun and that of object C is never more than 3 hours. Thus, it is an inner planet. The elongation of this planet can be found by calculating difference between the R.A. of the two. We note that on Jan 1, the difference in R. A. is e = 18h 44m7 s .11 − 15h 28m49s .55 = 3h 15m17s .56 ≈ 48◦ 49′ (12.78) Here the implicit assumption is the planet is exactly on the ecliptic and hence difference in R.A. will correctly give the elongation. Now in case of the Mercury, the orbital radius is about 0.4 A. U. Thus, its maximum elongation is will be about 28◦ . Orbital radius of the Venus is about 0.72 A.U. and the maximum elongation (assuming circular orbits) is about 47◦ . Thus, we note that elongation of object C is way more than maximum elongation of the Mercury but reasonably close to the maximum elongation of the Venus. The small difference can be purely due to our approximations (measuring only difference along R.A. and assuming circular orbits). We can safely conclude that the object C is the Venus. A clear sign of outer planet is the fact that during the course of the year, it can be seen sometime at the opposition. In the terms of coordinates, it means the difference between R.A. of the object and R.A. of the Sun should be 12 hours. Both object B (February 3) and object D (August 31) can be seen at opposition. Thus, they are outer planets. We next note that object D is changing coordinates much faster than object B. Let us concentrate on the object B first. These objects are changing coordinates from day to day. However, we must remember that this change is combined effect of the changing position of the Earth and changing position of the object. In usual cases, it will be difficult to separate the two contributions. However, as the table gives coordinates over the entire year, we can say that the orbital position of the Earth as on Jan. 1 will be nearly the same as that on Dec. 31. Thus difference in coordinates of the objects between these two dates, will be entirely because of changing position of the object. For object B, difference in R.A. over one year is nearly 2h i.e. it will take nearly 12 years to go through full circle (24h ). We can
12.2 Data Analysis 119 recall that synodic period of the Jupiter is 12 years and conclude that the object B is the Jupiter. As object D is moving faster than object D, we conclude the object D is the Mars. (b) To be visible for the longest duration at night, the object should rise at the sunset and set nearly at the sunrise. This again means that it should be at opposition. We have already found dates of opposition of the Jupiter and the Mars above. Next we note that the winter nights are longer than summer nights. As the observer is based in the northern hemisphere, we can say, February nights will be longer than August nights. Thus, we will conclude that the Jupiter will be seen for the longest duration at some night. (c) The corresponding date will be the date of opposition i.e. February 3. (d) To place all objects in the orbit diagram as on February 3, we have to take R.A. of each of these objects. As we are viewing the solar system perfectly face-on, declination is not relevant here. To place the earth, we take R.A. of the Sun. Orbit diagram shows direction of the vernal equinox. If the Earth was exactly opposite to that direction (lower part of the paper), observer will see R.A. of the Sun as 0h . The radial vector of the Earth currently makes e = 24h 0 m0 s − 21h 4 m47s .25 = 2h 55m12s .75 ≈ 44◦ (12.79) angle with the vertical line. When viewed from the top, the planets move anti-clockwise. As we are few days before vernal equinox, the Earth’s position will be to the left of the vertical line. The four orbits denote Venus, Earth, Mars and Jupiter. So we place the Earth making angle of 44◦ with the vertical line on the second (from inside) orbit on the lower left part. The Jupiter is at opposition on that date. So it will be placed in the last orbit exactly along the Sun-Earth line. Next we note that vernal equinox direction points to reference at infinity. Hence, direction of 0h as drawn from the Earth will also be exactly vertically up. From that direction, as we go anti-clockwise, the R.A. will go on increasing. We will place the Venus and the Mars in resepctive orbits as per their R.A.
120 Solutions: Celestial Coordinate Systems
Chapter 13 Solutions: Geometric Astronomy and Time 1. The apperant angular separation between the Sun and the Venus as seen from the Earth is maximum, when ∡SV E = 90◦ . See figure. ∴ dSV = d⊙−⊕ sin 46◦ (13.1) = (1A.U.) × 0.719 ≈ 0.72A.U. (13.2) 2. During a full Moon we see the whole face of the Moon. Hence angle in radians = diameter of moon distance to moon (13.3) ∴ distance to moon = diameter of moon angle in radians = 2 × 1.7374 × 106m (0.46π)/180 = 2 × 1.7374 × 106 × 180 0.46π = 4.3 × 108m (13.4) 3. From the very definition of unit parsec, angle in arc seconds = diameter of orbit in A.U. distance in parsec (13.5) = 2 100 = 0.02′′ (13.6)
122 Solutions: Geometric Astronomy and Time 4. By Kepler’s third Law, T 2 M T 2 E = a 3 M a 3 E (13.7) ∴ T 2 M = 1.523 × (365.2564)2 TM = 1.521.5 × 365.2564 TM = 684.48 days (13.8) To find time betwen successive oppositions, we have to find the synodic period of the Mars. 1 TSy = 1 TE − 1 TM (13.9) 1 TSy = 1 365.2564 − 1 684.48 TSy = 783.18 days (13.10) Thus the time between successive oppositions (synodic period) of Mars is 783 days. 5. First we find orbital period and synodic period of the Mars. TM = vuut R3 M R3 ⊕ T⊕ (13.11) = q (1.524)3 = 1.8814 years = 687.18 days (13.12) 1 Tsy = 1 T⊕ − 1 TM (13.13) = T⊕TM TM − T⊕ (13.14) = 1.8814 1.8814 − 1 years (13.15) = 779.65 days (13.16) This means there is an opposition of Mars about every 780 days. To find the dats of great opposition in the year 2018, 15 × 365 + 4 = 5479 days (13.17) 5479 779.65 = 7.0275 (13.18)
123 It means that there will be 7 oppositions between Aug. 28, 2003 and Aug. 28, 2018. So the date for the great opposition in 2018 should be 5479 − 7 × 779.65 = 21.38days (13.19) i.e. 21.4 days before Aug. 28, 2018. It is on Aug. 7, 2018. 6. The length of solar day is exactly 24 hrs. But sidereal day is a little less than solar day because of the yearly motion of the Sun. Now, in one solar year the Earth completes 365.25 solar days but it completes one additional rotation around itself. Thus number of sidereal days will be 366.25. Tsid = TSol × n n + 1 (13.20) = 24h × 365.25 365.25 + 1 = 23h 56m4 s (13.21) Now, if Earth rotates in opposite direction then exactly opposite thing will happen i.e. the Earth will complete one rotation less than number of solar days in a solar year. On the other hand, the Sidereal day will remain the same as it doesn’t depends on the rotation direction. (neglecting the precession) T ′ Sol = Tsid × n − 1 n (13.22) = TSol × n n + 1 × n − 1 n = 24h × 365.25 − 1 365.25 + 1 = 23h 52m8 s (13.23) 7. In the Figure 13.1, the observer is located at O and the satellite is at
124 Solutions: Geometric Astronomy and Time E R O Z R S S δ ϕ Zenith Figure 13.1 – Geocentric parallax of the satellite S. ∠EOS = 180 − z = 180 − 46.0 = 134◦ (13.24) δ = 180 − ϕ − ∡∠EOS = 10.4 ◦ (13.25) Rs sin(∡EOS) = R⊕ sin δ (13.26) Rs R⊕ = sin(∡EOS) sin δ = sin 134◦ sin 10.4 ◦ ∴ Rs = 3.98R⊕ (13.27) The angle δ is called geocentric parallax of any near-Earth object, such as the satellite. 8. If the observer was at the centre of the Mars (C), Phobos would have been visible above horizon for exactly half the period. However, the observer is on martian surface (O), thus as shown in the figure 13.2, the Phobos will set at point P.
125 R Ph M θ O C horizon P a Figure 13.2 – Horizon for martian observer Tph = 2π vuut a 3 ph GMMars (13.28) = 2π s (9.38 × 106 ) 3 6.6726 × 10−11 × 6.421 × 1023 = 2.76 × 104 sec Tph ≈ 7.660hrs (13.29) 1 Tsyn = 1 Tph − 1 TMars (13.30) Tsyn = TMarsTph TMars − Tph = 24.623 × 7.660 24.623 − 7.660 = 11.12hrs (13.31)
126 Solutions: Geometric Astronomy and Time S M2 M1 E θ Figure 13.3 – Mars at the opposition and at the quadrature This is synodic period of Phobos for a martian observer. Now, θ = cos−1 RMars RP h (13.32) t = 2θ 360◦ × Tsyn = Tsyn 180◦ cos−1 RMars RP h (13.33) = 11.12 180 cos−1 3393 9380 = 4.25hrs (13.34) Hence Phobos will be visible in the martian sky for about 4 hours and 15 minutes 9. If we go into the reference frame of the Earth, the Mars will complete one revolution in its synodic period. In the figure 13.3, length SM2 is orbital radius of the Mars (am) and SE is orbital radius of the Earth
127 (a⊕). Angle between the two positions is θ. 1 Tsy = 1 T⊕ − 1 Tm (13.35) ∴ Tsy = TmT⊕ (Tm − T⊕) (13.36) = 687 × 365 (687 − 365) ∴ Tsy = 778.7days (13.37) θ = t Tsy × 360◦ = 106 779 × 360◦ (13.38) = 49.00◦ (13.39) a⊕ = am cos θ (13.40) am = 1 cos 49.00◦ am = 1.52A.U. (13.41) 10. (a) Drawing scaled diagram is impossible. Rough sketch is accepted. (b) There are 10 days and nights per taris year. The obliquity is 3◦ , which means that the planet’s rotation is in the same direction as its orbit. Thus, total number of rotations per year is 10 + 1 = 11. Note: The obliquity is positive (similar to the Earth / Mars / Jupiter). This means, we have ADD one rotation. Subtracting one rotation by assuming opposite rotation (like the Venus) is not correct. (c) By Kepler’s third law, T 2 en R3 en = T 2 ex R3 ex (13.42) R 3 ex = 1.6 2R3 en 0.2 2 Rex = √3 64 = 4 endorlengths (13.43)
128 Solutions: Geometric Astronomy and Time T C c b a f e d B A S (d) Using same logic as above T 2 C R3 C = T 2 T R3 T (13.44) T 2 C = 9 3R3 T T 2 T R3 T TC = √ 729 = 27 tarisyears (13.45) (e) As Corulus is in Opposition, Sola - Taris - Corulus form straight line (in that order). Distance = 9 - 1 = 8 tarislengths. (f) In the figure 10f, S is Sola, A and B are start of the year positions of Taris and Corulus, T and C are their positions after ’n’ days. Angles are named from a to f. The dashed line is parallel to line SB. Triangle(SCT) is used for sine rule as well as answer in the next part. Figure is not to the scale.
129 a + b + c = π (13.46) b + d + e = π (13.47) d = f + c (13.48) f + c = 2πn 10 (13.49) f = 2πn 270 (13.50) sin b = 9 sin a (13.51) e = π − b − d (13.52) = π − b − c − f e = a − f (13.53) b = π − (a + c) (13.54) = π − a + 2πn 10 − 2πn 270 b = π − a + 52πn 270 (13.55) 9 sin a = sin π − a + 52πn 270 (13.56) = sin a + 52πn 270 = sin a cos 52πn 270 + cos a sin 52πn 270 (13.57) 9 = cos 52πn 270 + cot a sin 52πn 270 cot a = 9 − cos 52πn 270 sin 52πn 270 (13.58) a = tan−1 sin 52πn 270 9 − cos 52πn 270 (13.59) λ = π − e (13.60) = π + f − a λ = π + 2πn 270 − tan−1 sin 52πn 270 9 − cos 52πn 270 (13.61)
130 Solutions: Geometric Astronomy and Time (g) When n = 1 a = tan−1 sin 52π 270 9 − cos 52π 270 (13.62) = tan−1 sin 0.605 9 − cos 0.605 = 0.0756 rad = 4◦ 20′ (13.63) sin b = 9 sin a (13.64) = 9 sin(4◦ 20′ ) = 0.680 (13.65) Area = 1 2 × l(ST) × l(SC) × sin b (13.66) = 1 2 × 1 × 9 × 0.680 = 3.06 (13.67) The area is about 3 (tarislength) 2 . 11. Since the observer is close to the pole, the affect of the earth’s rotation on the transit could be neglected. The Sun’s angular size for the observer will be, θ⊙ = 2R⊙ 1AU (13.68) = 2 × 6.955 × 108 1.496 × 1011 ! × 206265 60 θ⊙ ≈ 32.0 ′ (13.69) θV enus = 2RV enus dV −E (13.70) = 2 × 0.949 × 6.371 × 106 × 206265 (1 − 0.723) × 1.496 × 1011 × 60 θV enus ≈ 1.00′ (13.71) The angular velocity of the Venus around the Sun, with respect to the earth will be, ωsy = ωV enus − ωEarth = 2π TV enus − 2π TEarth (13.72) = 2π × 206265 86400 × 60 ( 1 224.70 − 1 365.25 ) ≈ 4.28 × 10−4 arcmin/s (13.73)
131 Figure 13.4 – Venus Transit Geometry For the observer on earth, let us say that the Venus moved through angle θ during the whole transit. Let OE be perpendicular to AB, OA = θ⊙ ∡AOB = 90◦ MN k AB ∴ OE = θ⊙ sin 45◦ = 11.3 ′ (13.74) OC = (θV enus + θ⊙) 2 = 16.5 ′ (13.75) CE = √ OC2 − OE2 (13.76) = √ 16.0 2 − 11.3 2 ≈ 12.0 ′ (13.77) CD = 2CE ∴ CD = ∡CFD = θ = 24.0 ′ (13.78) However CFD is the angle measured from the Earth, which is different
132 Solutions: Geometric Astronomy and Time from the angle measured from the Sun. Angle measured from the Sun (θ ′ = ∡COD) is the actual angle that Venus covered during the transit. sin(θ/2) sin(θ ′/2) = aV enus dVE (13.79) = 0.723 (1 − 0.723) θ ′ = 2 sin−1 sin 12′ 2.610 ! = 9.195′ (13.80) ttransit = θ ′ ωsy (13.81) = 9.195′ 4.28 × 10−4 ttransit = 5h 58m (13.82) So the transit will finish at about 14h58m. 12. Let us call the semi-major axis of Moon’s orbit as a, its eccentricity as e, the revolution period T, apparent radius r, the distance between Earth and Moon as d, the apparent radius of the Sun as r⊙. When the Moon is at perigee, the total eclipse will be longest and when the Moon is at apogee, the annular eclipse will be longest. By conserving angular momentum, d 2 apωap = d 2 periωperi (13.83) ∴ ωperi ωap = a(1 + e) a(1 − e) !2 = (1 + e) 2 (1 − e) 2 (13.84) As the Earth’s orbit around the Sun is circular, we can assume ω⊕ is constant. Thus, time of eclipse at these two points, tap = 2(r⊙ − rap) ωap (13.85) tperi = 2(rperi − r⊙) ωperi (13.86) ∴ tap tperi = ωperi ωap 2(r⊙ − rap) 2(rperi − r⊙) = (1 + e) 2 (1 − e) 2 (r⊙ − rap) (rperi − r⊙) (13.87)